I'm trying to show that the following statement is true:
If $||\mathbf a - \theta \mathbf b||^2 - ||\mathbf a||^2 \geq 0$ for all $\theta \in [0,1]$, then $\mathbf a^T \mathbf b \le 0$.
Is this how I would go about this?? $$\sqrt{a - \theta b}^2 - \sqrt{a}^2 \geq 0$$
$$a- \theta b - a \geq 0$$
I'm not sure how to get to the right conclusion.
(I'm using the Euclidean Norm)
$$ || a - \theta b||^2 - || a||^2 = || a||^2 + \theta^ 2 || b||^2 - 2\theta a\cdot b - || a||^2 = \theta^ 2 || b||^2 - 2\theta a\cdot b $$
this function of theta is minimum when $$ \theta = \frac{a\cdot b}{|| b||^2} $$
If $ a\cdot b > 0 $ this value is in $[0,1]$.
$$ 0\le f\left(\frac{a\cdot b}{|| b||^2}\right) = -\frac {(a\cdot b)^2}{|| b||^2} $$
This is impossible.